Question

If $A+B+C=\pi $, prove that $\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\cos C$.

Answer 1

Hint: For solving this question we will use some trigonometric formula like formula for $\cos C+\cos D$, $\cos C-\cos D$ and $\cos 2\theta $ for simplifying the term written on the left-hand side. After that, we will prove it equal to the term on the right-hand side.

Complete step-by-step answer:

Given:

It is given that if $A+B+C=\pi $ and we have to prove the following equation:

$\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\cos C$

Now, before we proceed we should know the following four formulas:

$\begin{align}

  & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)....................\left( 1 \right) \\

 & \cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)...................\left( 2 \right) \\

 & \cos 2\theta =2{{\cos }^{2}}\theta -1.........................................................\left( 3 \right) \\
 & A+B+C=\pi \\

 & \Rightarrow A+B=\pi -C \\

 & \Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right)=-\cos C..............................\left( 4 \right) \\

\end{align}$

Now, we will be using the above four formulas to simplify the term on the left-hand side to prove the desired result.

Now, L.H.S is equal to $\cos 2A+\cos 2B-\cos 2C$ so, using the formula from equation (1).
Then,

$\begin{align}

  & \cos 2A+\cos 2B-\cos 2C \\

 & \Rightarrow 2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)-\cos 2C \\

 & \Rightarrow 2\cos \left( A+B \right)\cos \left( A-B \right)-\cos 2C \\

\end{align}$

Now, using the formula from equation (3) and equation (4) in the above equation. Then,
$\begin{align}

  & 2\cos \left( A+B \right)\cos \left( A-B \right)-\cos 2C \\

 & \Rightarrow -2\cos C\cos \left( A-B \right)-\left( 2{{\cos }^{2}}C-1 \right) \\

 & \Rightarrow -2\cos C\cos \left( A-B \right)-2{{\cos }^{2}}C+1 \\

\end{align}$

$\begin{align}

  & \Rightarrow 1+2\cos C\left( -\cos \left( A-B \right)-\cos C \right) \\

 & \Rightarrow 1+2\cos C\left( \cos \left( A+B \right)-\cos \left( A-B \right) \right) \\

\end{align}$

Now, using the formula from the equation (2) in the above equation. Then,

$\begin{align}

  & 1+2\cos C\left( \cos \left( A+B \right)-\cos \left( A-B \right) \right) \\

 & \Rightarrow 1+2\cos C\left( -2\sin \left( \dfrac{A+B+A-B}{2} \right)\sin \left( \dfrac{A+B-A+B}{2} \right) \right) \\

 & \Rightarrow 1+2\cos C\left( -2\sin A\sin B \right) \\

 & \Rightarrow 1-4\sin A\sin B\cos C \\

\end{align}$

Now, from the above result, we can say that $\cos 2A+\cos 2B-\cos 2C=1-4\sin A\sin B\cos C$.

Thus, $L.H.S=R.H.S$

Hence Proved.

Note: Here, the student should first understand what we have to prove in the question and then proceed in a stepwise manner while solving. For making the simplification easier, we should also try to make use of trigonometric results like $\cos \left( \pi -\theta \right)=-\cos \theta $ for making equations that will help us further in the solution. Moreover, the formulas like $\cos C+\cos D$ , $\cos C-\cos D$ and $\cos 2\theta $ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.